Question
Hassan wants to add a border to a photo he took before he frames it. The final area of the entire framed picture will be 96 square inches. The length of the picture is 10 inches, and the width is 6 inches. Which of the following is the width of the border?(1 point)
Responses
3 inches
9 inches
1 inch
4 inches
May’s class is testing their egg protection contraptions by dropping them off the roof of their school. Use GeoGebra to graph the situation using the formula y=−16t^2+40
. Which of the following correctly interprets the height of the school’s roof?(1 point)
Responses
41 feet
15 feet
40 feet
20 feet
Luca is in a culvert below street level. He launches an object at an upward velocity of 40 feet per second. Use GeoGebra to graph the situation using the formula y=−16t^2+40t−5
. How far above or below street level is Luca’s initial launching point?(1 point)
Responses
5 feet above
2.5 feet below
0 feet above or below
5 feet below
Which of the following scenarios would use a quadratic equation to solve?(1 point)
Responses
A coach models the trajectory of a ball to find out how high the player is kicking the ball.
An accountant models the growth of a savings account that doubles every month to find how much will be in the account in 10 years.
The trajectory of a balloon that rises 2 feet per second is modeled to find how high the balloon will be in 5 minutes.
A child places a quarter into a piggy bank every week.
Responses
3 inches
9 inches
1 inch
4 inches
May’s class is testing their egg protection contraptions by dropping them off the roof of their school. Use GeoGebra to graph the situation using the formula y=−16t^2+40
. Which of the following correctly interprets the height of the school’s roof?(1 point)
Responses
41 feet
15 feet
40 feet
20 feet
Luca is in a culvert below street level. He launches an object at an upward velocity of 40 feet per second. Use GeoGebra to graph the situation using the formula y=−16t^2+40t−5
. How far above or below street level is Luca’s initial launching point?(1 point)
Responses
5 feet above
2.5 feet below
0 feet above or below
5 feet below
Which of the following scenarios would use a quadratic equation to solve?(1 point)
Responses
A coach models the trajectory of a ball to find out how high the player is kicking the ball.
An accountant models the growth of a savings account that doubles every month to find how much will be in the account in 10 years.
The trajectory of a balloon that rises 2 feet per second is modeled to find how high the balloon will be in 5 minutes.
A child places a quarter into a piggy bank every week.
Answers
GPT-4o mini
Let's go through each question one by one.
1. **Hassan's Photo Border Width**:
- The area of the framed picture is given as 96 square inches.
- The dimensions of the picture are 10 inches (length) and 6 inches (width).
- The area of the picture itself is \(10 \times 6 = 60\) square inches.
- If we let \(x\) be the width of the border, the dimensions of the framed picture will be \((10 + 2x)\) by \((6 + 2x)\).
- The area of the framed picture will then be:
\[
(10 + 2x)(6 + 2x) = 96
\]
- Expanding this:
\[
60 + 20x + 12x + 4x^2 = 96
\]
\[
4x^2 + 32x + 60 - 96 = 0
\]
\[
4x^2 + 32x - 36 = 0
\]
Dividing the entire equation by 4:
\[
x^2 + 8x - 9 = 0
\]
To solve for \(x\), we can use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-8 \pm \sqrt{8^2 - 4(1)(-9)}}{2(1)} = \frac{-8 \pm \sqrt{64 + 36}}{2} = \frac{-8 \pm 10}{2}
\]
This gives us:
\[
x = 1 \quad \text{(Positive solution)}
\]
or
\[
x = -9 \quad \text{(Not applicable)}
\]
- **The width of the border is 1 inch.**
2. **Height of the School’s Roof**:
- The formula given is \(y = -16t^2 + 40\).
- This is a quadratic equation representing the height of an object over time, with the height at \(t = 0\) being \(y = -16(0)^2 + 40 = 40\) feet.
- Therefore, **the height of the school's roof is 40 feet.**
3. **Luca’s Launching Point**:
- The formula is \(y = -16t^2 + 40t - 5\).
- To find the initial launching point, we evaluate at \(t = 0\):
\[
y = -16(0)^2 + 40(0) - 5 = -5
\]
- Therefore, **Luca’s initial launching point is 5 feet below street level.**
4. **Scenario Using a Quadratic Equation**:
- Let's evaluate each scenario:
- **Scenario 1**: A coach models the trajectory of a ball to find out how high the player is kicking the ball. (This scenario uses a quadratic equation because it involves projectile motion.)
- **Scenario 2**: An accountant models the growth of a savings account that doubles every month. (This is exponential growth, not quadratic.)
- **Scenario 3**: The trajectory of a balloon that rises 2 feet per second. (This is linear, not quadratic.)
- **Scenario 4**: A child places a quarter into a piggy bank every week. (This is also linear.)
- Therefore, **the scenario that uses a quadratic equation is the first one** about the coach modeling the trajectory of the ball.
**Final Answers**:
1. 1 inch
2. 40 feet
3. 5 feet below street level
4. A coach models the trajectory of a ball to find out how high the player is kicking the ball.
1. **Hassan's Photo Border Width**:
- The area of the framed picture is given as 96 square inches.
- The dimensions of the picture are 10 inches (length) and 6 inches (width).
- The area of the picture itself is \(10 \times 6 = 60\) square inches.
- If we let \(x\) be the width of the border, the dimensions of the framed picture will be \((10 + 2x)\) by \((6 + 2x)\).
- The area of the framed picture will then be:
\[
(10 + 2x)(6 + 2x) = 96
\]
- Expanding this:
\[
60 + 20x + 12x + 4x^2 = 96
\]
\[
4x^2 + 32x + 60 - 96 = 0
\]
\[
4x^2 + 32x - 36 = 0
\]
Dividing the entire equation by 4:
\[
x^2 + 8x - 9 = 0
\]
To solve for \(x\), we can use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-8 \pm \sqrt{8^2 - 4(1)(-9)}}{2(1)} = \frac{-8 \pm \sqrt{64 + 36}}{2} = \frac{-8 \pm 10}{2}
\]
This gives us:
\[
x = 1 \quad \text{(Positive solution)}
\]
or
\[
x = -9 \quad \text{(Not applicable)}
\]
- **The width of the border is 1 inch.**
2. **Height of the School’s Roof**:
- The formula given is \(y = -16t^2 + 40\).
- This is a quadratic equation representing the height of an object over time, with the height at \(t = 0\) being \(y = -16(0)^2 + 40 = 40\) feet.
- Therefore, **the height of the school's roof is 40 feet.**
3. **Luca’s Launching Point**:
- The formula is \(y = -16t^2 + 40t - 5\).
- To find the initial launching point, we evaluate at \(t = 0\):
\[
y = -16(0)^2 + 40(0) - 5 = -5
\]
- Therefore, **Luca’s initial launching point is 5 feet below street level.**
4. **Scenario Using a Quadratic Equation**:
- Let's evaluate each scenario:
- **Scenario 1**: A coach models the trajectory of a ball to find out how high the player is kicking the ball. (This scenario uses a quadratic equation because it involves projectile motion.)
- **Scenario 2**: An accountant models the growth of a savings account that doubles every month. (This is exponential growth, not quadratic.)
- **Scenario 3**: The trajectory of a balloon that rises 2 feet per second. (This is linear, not quadratic.)
- **Scenario 4**: A child places a quarter into a piggy bank every week. (This is also linear.)
- Therefore, **the scenario that uses a quadratic equation is the first one** about the coach modeling the trajectory of the ball.
**Final Answers**:
1. 1 inch
2. 40 feet
3. 5 feet below street level
4. A coach models the trajectory of a ball to find out how high the player is kicking the ball.